A tire that has been run flat has just been removed from an Alcoa aluminum rim. Technician A states to roll the rim on a flat le
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Answer:
x=±0.026m
Explanation:
In simple harmonic motion the maximum value of the magnitude of velocity
The speed as a function of position for simple harmonic oscillator is given by
where A is amplitude of motion
Given data
Amplitude A=3 cm =0.03 m
v=(1/2)Vmax
To find
We have asked to find position x does its speed equal half of is maximum speed
Solution
The speed of the particle the maximum speed as:
x=±(√3(0.03)/2)
x=±0.026m
Answer:
The acceleration of the box is 2.05 m/s²
Explanation:
The given parameters of the motion of the box are;
The mass of the box, m = 50 kg
The pulling force, acting on the box = 200 N
The angle at which the force acts, θ = 30° above the horizontal
The coefficient of kinetic friction, = 0.25
The normal reaction from the box resting on a flat surface, N = The weight of the box, W - The vertical component of the pulling force,
N = W - = m·g -
× sin(θ)
Where;
g = The acceleration due to gravity = 9.8 m/s²
∴ N = W - = m·g -
× sin(θ) = 50 kg × 9.8 m/s² - 200 N × sin(30°)
∴ N = 490 N - 200 N × 0.5 = 390 N
The normal reaction, N = 390 N
The force of friction, = The coefficient of kinetic friction,
× The normal reaction, N
∴ =
× N = 0.25 × 390 N = 97.5 N
The net force, , acting on the block = The pulling force,
- The friction force,
∴ =
-
= 200 N - 97.5 N = 102.5 N
= 102.5 N
According to Newton's second law of motion on the net force acting on an object, we have;
= m × a
Where;
a = The acceleration of the box
∴ a = /m = 102.5 N/(50 kg) = 2.05 m/s²
The acceleration of the box = a = 2.05 m/s².